Chemical Kinetics Class 12: Concepts, Formulas and Board Numericals

What Chemical Kinetics is about

Weightage: This chapter is Unit III, Chemical Kinetics, which carries 7 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit III Chemical Kinetics

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper. See the questions

Key concepts

Rate of a reaction: average and instantaneous

NCERT section 3.1 Rate of a Chemical Reaction

The rate of a reaction is the change in concentration of a reactant or product per unit time. For a reaction R → P:
$r_{av}=-\frac{\Delta [R]}{\Delta t}=+\frac{\Delta [P]}{\Delta t}$
The minus sign makes the rate positive, because $[R]$ decreases. Units are mol L⁻¹ s⁻¹ (or atm s⁻¹ for gases).

The average rate depends on the interval chosen and falls as the reaction proceeds. The instantaneous rate is the rate at one moment: the limit as $\Delta t\to 0$, found from the slope of the tangent to the concentration–time curve at that instant: $r_{inst}=-\frac{d[R]}{dt}$.

When coefficients are unequal, divide each rate of change by its coefficient so that the rate of the reaction has one value. For $aA+bB\to cC+dD$:
$\text{Rate}=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{c}\frac{d[C]}{dt}=\frac{1}{d}\frac{d[D]}{dt}$

Worked example

For the reaction $2N_2{O}_5(g)\to 4N{O}_2(g)+O_2(g)$, N₂O₅ disappears at $4.0\times 10^{-4}$ mol L⁻¹ s⁻¹ at one instant. Find the rate of the reaction and the rates of formation of NO₂ and O₂. (2 marks)

Try it first, then show the solutionSolution

1. Rate $=-\frac{1}{2}\frac{d[N_2{O}_5]}{dt}=\frac{1}{4}\frac{d[N{O}_2]}{dt}=\frac{d[O_2]}{dt}$.
2. Rate of reaction $=\frac{1}{2}\times 4.0\times 10^{-4}=2.0\times 10^{-4}$ mol L⁻¹ s⁻¹. (1)
3. Formation of NO₂ $=4\times 2.0\times 10^{-4}=8.0\times 10^{-4}$; formation of O₂ $=2.0\times 10^{-4}$ mol L⁻¹ s⁻¹. (1)

Answer: Rate = 2.0 × 10⁻⁴; NO₂ forms at 8.0 × 10⁻⁴ and O₂ at 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹.

Rate law, order and the rate constant

NCERT section 3.2 Factors Influencing Rate of a Reaction (3.2.1 Dependence of Rate on Concentration; 3.2.2 Rate Expression and Rate Constant; 3.2.3 Order of a Reaction)

Rate depends on concentration, temperature and the presence of a catalyst. The dependence on concentration is written as a rate law (rate expression), found by experiment:
$\text{Rate}=k[A{]}^x[B{]}^y$
The powers $x$ and $y$ need not equal the coefficients in the balanced equation. $k$ is the rate constant (the rate when every concentration is 1 mol L⁻¹).

The order is the sum $x+y$. It can be 0, 1, 2, 3 or even a fraction. Order with respect to one reactant is its own power.

Units of k for order $n$: $({\text{mol L}}^{-1}{)}^{1-n}{\text{s}}^{-1}$

• zero order: mol L⁻¹ s⁻¹
• first order: s⁻¹
• second order: L mol⁻¹ s⁻¹

Initial-rate method: change one concentration at a time and see how the rate changes. If doubling [A] doubles the rate, the order in A is 1; if it makes the rate four times, the order is 2; if the rate does not change, the order is 0.

Worked example

For the reaction A + B → products, the following initial rates were measured:
Experiment 1: [A] = 0.1, [B] = 0.1 mol L⁻¹, rate = 2.0 × 10⁻³ mol L⁻¹ s⁻¹
Experiment 2: [A] = 0.2, [B] = 0.1, rate = 8.0 × 10⁻³
Experiment 3: [A] = 0.1, [B] = 0.2, rate = 4.0 × 10⁻³
Find the order with respect to A and B, write the rate law and calculate k. (3 marks)

Try it first, then show the solutionSolution

1. Experiments 1 and 2: [A] doubles, rate goes from 2.0 to 8.0 (×4), so order in A = 2. (½)
2. Experiments 1 and 3: [B] doubles, rate doubles, so order in B = 1. (½)
3. Rate $=k[A{]}^2[B]$; overall order 3. (1)
4. $k=\frac{2.0\times 10^{-3}}{(0.1{)}^2\times 0.1}=2.0$ L² mol⁻² s⁻¹. (1)

Answer: Second order in A, first order in B; Rate = k[A]²[B]; k = 2.0 L² mol⁻² s⁻¹.

Molecularity, elementary reactions and mechanism

NCERT section 3.2.4 Molecularity of a Reaction

An elementary reaction takes place in one step. Many reactions are complex: they occur through a series of elementary steps called the mechanism.

Molecularity is the number of reacting species (atoms, ions or molecules) that must collide at the same time in an elementary step. It is unimolecular (NH₄NO₂ → N₂ + 2H₂O), bimolecular (2HI → H₂ + I₂) or, rarely, termolecular. More than three particles colliding at once is very unlikely.

Order versus molecularity:

• Order is found by experiment; molecularity is a theoretical idea for an elementary step.
• Order can be zero or a fraction; molecularity is always a whole number (1, 2 or 3), never zero.
• Order applies to both elementary and complex reactions; molecularity has no meaning for a complex reaction as a whole.

In a complex reaction, the overall rate is controlled by the slowest step, the rate-determining step. The decomposition of H₂O₂ catalysed by I⁻ in alkaline medium goes in two steps:
(1) $H_2{O}_2+I^{-}\to H_{2}O+I{O}^{-}$ (slow)
(2) $H_2{O}_2+I{O}^{-}\to H_{2}O+I^{-}+O_2$ (fast)
The slow step makes the rate law $\text{Rate}=k[H_2{O}_2][I^{-}]$: first order in each, second order overall. IO⁻ is an intermediate.

Worked example

For the iodide-catalysed decomposition of H₂O₂ described above: (a) write the rate law, (b) give the order and the molecularity of the slow step, (c) identify the intermediate and the catalyst. (3 marks)

Try it first, then show the solutionSolution

1. (a) Rate = k[H₂O₂][I⁻], from the slow step. (1)
2. (b) Overall order 2 (1 in H₂O₂, 1 in I⁻); the slow step is bimolecular. (1)
3. (c) IO⁻ is formed in step 1 and used in step 2: intermediate. I⁻ is used in step 1 and returned in step 2: catalyst. (1)

Answer: Rate = k[H₂O₂][I⁻]; order 2, slow step bimolecular; IO⁻ is the intermediate and I⁻ the catalyst.

Zero order reactions

NCERT section 3.3.1 Zero Order Reactions

In a zero order reaction the rate does not depend on the concentration of the reactant: $\text{Rate}=k[R{]}^0=k$.

Integrating $-\frac{d[R]}{dt}=k$ gives
$[R]=[R{]}_0-kt$, so $k=\frac{[R{]}_0-[R]}{t}$
A plot of [R] against t is a straight line with slope $-k$ and intercept $[R{]}_0$.

Half-life: put $[R]=[R{]}_0/2$:
$t_{1/2}=\frac{[R{]}_0}{2k}$
It is directly proportional to the initial concentration.

Zero order reactions are uncommon and occur under special conditions, such as on the surface of a catalyst that is fully covered. Examples in NCERT: decomposition of ammonia on a hot platinum surface at high pressure ($2N{H}_3(g)\to N_2(g)+3H_2(g)$), and thermal decomposition of HI on a gold surface.

Worked example

A zero order reaction has k = 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹ and [R]₀ = 0.10 mol L⁻¹. (a) How long does it take for [R] to fall to 0.04 mol L⁻¹? (b) What is its half-life? (2 marks)

Try it first, then show the solutionSolution

1. (a) $t=\frac{[R{]}_0-[R]}{k}=\frac{0.10-0.04}{2.0\times 10^{-4}}=300$ s. (1)
2. (b) $t_{1/2}=\frac{[R{]}_0}{2k}=\frac{0.10}{4.0\times 10^{-4}}=250$ s. (1)

Answer: (a) 300 s (b) 250 s

First order reactions and half-life

NCERT section 3.3.2 First Order Reactions; 3.3.3 Half-Life of a Reaction

In a first order reaction, $\text{Rate}=k[R]$. Integrating gives
$\ln [R]=-kt+\ln [R{]}_0$, or $k=\frac{2.303}{t}\log \frac{[R{]}_0}{[R]}$

Graphs: $\ln [R]$ against $t$ is a straight line with slope $-k$; $\log \frac{[R{]}_0}{[R]}$ against $t$ is a straight line through the origin with slope $\frac{k}{2.303}$.

Half-life: put $[R]=[R{]}_0/2$:
$t_{1/2}=\frac{0.693}{k}$
It does not depend on the initial concentration. So each half-life halves what is left: after $n$ half-lives, $\frac{1}{2^n}$ remains.

Useful results: time for 75% reaction = 2 half-lives; time for 99.9% reaction ≈ 10 half-lives.

For a gas-phase first order reaction A(g) → B(g) + C(g) followed by total pressure: $p_A=2p_i-p_t$, so
$k=\frac{2.303}{t}\log \frac{p_i}{2p_i-p_t}$
Examples: hydrogenation of ethene, decomposition of N₂O₅ and N₂O.

Pseudo first order reactions: when one reactant is present in large excess, its concentration stays almost constant and the reaction behaves as first order in the other. Examples: acid hydrolysis of ethyl acetate in excess water, and inversion of cane sugar, $\text{Rate}=k[C_{12}{H}_{22}{O}_{11}]$.

Worked example

A first order reaction has k = 0.0693 min⁻¹. Calculate (a) its half-life, (b) the time for 75% completion, and (c) the time for 90% completion. (3 marks)

Try it first, then show the solutionSolution

1. (a) $t_{1/2}=\frac{0.693}{0.0693}=10$ min. (1)
2. (b) 75% done means ¼ left = two half-lives: 20 min. Check: $t=\frac{2.303}{0.0693}\log 4=33.23\times 0.602=20$ min. (1)
3. (c) $t=\frac{2.303}{0.0693}\log \frac{100}{10}=33.2\times 1=33.2$ min. (1)

Answer: (a) 10 min (b) 20 min (c) about 33.2 min

Temperature dependence: the Arrhenius equation

NCERT section 3.4 Temperature Dependence of the Rate of a Reaction

For most reactions, a 10 degree rise in temperature nearly doubles the rate constant. The Arrhenius equation expresses this:
$k=A{e}^{-E_a/RT}$
$A$ is the frequency (pre-exponential) factor and $E_a$ is the activation energy, in J mol⁻¹.

Reactants must pass through an unstable activated complex at the top of the energy barrier. $E_a$ is the extra energy (above the average) that colliding molecules need to reach it. The term $e^{-E_a/RT}$ is the fraction of molecules with kinetic energy equal to or greater than $E_a$. Raising the temperature shifts the Maxwell–Boltzmann distribution to higher energies and broadens it; the fraction above $E_a$ roughly doubles for a 10 degree rise.

Taking logs: $\ln k=\ln A-\frac{E_a}{RT}$. A plot of $\ln k$ against $1/T$ is a straight line with slope $-\frac{E_a}{R}$.

For two temperatures:
$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
Use R = 8.314 J mol⁻¹ K⁻¹ and temperatures in kelvin.

Worked example

The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Calculate the activation energy. (R = 8.314 J mol⁻¹ K⁻¹, log 2 = 0.3010) (3 marks)

Try it first, then show the solutionSolution

1. $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$ (1)
2. $0.3010=\frac{E_a}{2.303\times 8.314}\left(\frac{10}{300\times 310}\right)$ (1)
3. $E_a=0.3010\times 2.303\times 8.314\times 9300=53600$ J mol⁻¹ $\approx 53.6$ kJ mol⁻¹. (1)

Answer: Ea ≈ 53.6 kJ mol⁻¹

Effect of a catalyst and collision theory

NCERT section 3.4.1 Effect of Catalyst; 3.5 Collision Theory of Chemical Reactions

A catalyst increases the rate without being used up. It provides an alternative pathway (through an intermediate) with a lower activation energy, so more molecules have enough energy to react. A catalyst:

• does not change the Gibbs energy change (ΔG) of the reaction;
• catalyses only spontaneous reactions and cannot make a non-spontaneous one happen;
• does not change the equilibrium constant; it helps equilibrium to be reached faster, by speeding up the forward and backward reactions equally.

Collision theory treats molecules as hard spheres. A reaction happens only through collisions; the number of collisions per second per unit volume is the collision frequency, $Z$. But not every collision works. An effective collision needs:
1. enough energy (at least the threshold energy, i.e. average energy plus $E_a$), and
2. proper orientation, so that the right bonds can break and form.

The rate is written $\text{Rate}=P{Z}_{AB}{e}^{-E_a/RT}$, where $P$ is the probability or steric factor for proper orientation. For example, in the reaction of CH₃Br with OH⁻, only an OH⁻ approaching the side opposite to Br gives methanol.

Collision theory works well for simple molecules but less well for complex ones.

Worked example

A catalyst lowers the activation energy of a reaction by 20 kJ mol⁻¹ at 300 K. By roughly what factor does the rate constant increase, assuming A is unchanged? (R = 8.314 J mol⁻¹ K⁻¹; antilog 3.48 ≈ 3.0 × 10³) (3 marks)

Try it first, then show the solutionSolution

1. $\frac{k_{cat}}{k}=e^{(E_a-E_a^{\prime })/RT}$, so $\log \frac{k_{cat}}{k}=\frac{\Delta E_a}{2.303RT}$. (1)
2. $=\frac{20000}{2.303\times 8.314\times 300}=\frac{20000}{5744}=3.48$. (1)
3. $\frac{k_{cat}}{k}=\text{antilog }3.48\approx 3.0\times 10^3$. (1)

Answer: About 3000 times faster.

Chemistry in real life: Chemical Kinetics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Rate Law, Order and Molecularity
• Integrated Rate Equations and Half-Life
• Arrhenius Equation and Activation Energy

Chemical Kinetics formula sheet

Average and instantaneous rate

$r_{av}=-\frac{\Delta [R]}{\Delta t}=\frac{\Delta [P]}{\Delta t},r_{inst}=-\frac{d[R]}{dt}$

Units: mol L⁻¹ s⁻¹. Instantaneous rate = slope of the tangent.

Rate with stoichiometric coefficients

$\text{Rate}=-\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{c}\frac{d[C]}{dt}$

For aA + bB → cC + dD.

Rate law and order

$\text{Rate}=k[A{]}^x[B{]}^y,\text{order}=x+y$

x and y are found by experiment.

Units of the rate constant

$k:({\text{mol L}}^{-1}{)}^{1-n}{\text{s}}^{-1}$

Zero order mol L⁻¹ s⁻¹; first order s⁻¹; second order L mol⁻¹ s⁻¹.

Zero order: integrated rate equation

$[R]=[R{]}_0-kt,t_{1/2}=\frac{[R{]}_0}{2k}$

[R] against t: straight line, slope −k.

First order: integrated rate equation

$k=\frac{2.303}{t}\log \frac{[R{]}_0}{[R]}$

Also ln[R] = −kt + ln[R]₀ and [R] = [R]₀e^(−kt).

First order half-life

$t_{1/2}=\frac{0.693}{k}$

Independent of [R]₀. After n half-lives, (1/2)ⁿ is left.

Gas-phase first order reaction

$k=\frac{2.303}{t}\log \frac{p_i}{2p_i-p_t}$

For A(g) → B(g) + C(g), followed by total pressure p_t.

Useful first order times

$t_{75\%}=2t_{1/2},t_{99.9\%}\approx 10t_{1/2}$

From log 4 and log 1000.

Arrhenius equation

$k=A{e}^{-E_a/RT},\ln k=\ln A-\frac{E_a}{RT}$

ln k against 1/T: slope −Ea/R.

Arrhenius equation, two temperatures

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

R = 8.314 J mol⁻¹ K⁻¹; T in kelvin; Ea in J mol⁻¹.

Collision theory

$\text{Rate}=P{Z}_{AB}{e}^{-E_a/RT}$

Z: collision frequency; P: probability (steric) factor for proper orientation.

H₂O₂ decomposition (overall)

$2H_2{O}_2\to 2H_{2}O+O_2$

In alkaline medium with I⁻ catalyst; rate = k[H₂O₂][I⁻].

Zero order example

$2N{H}_3(g)\stackrel{1130K,Pt}{\to }{N}_2(g)+3H_2(g)$

On a platinum surface at high pressure; rate = k.

Chemical Kinetics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Chemical Kinetics

The question patterns to expect, and the method that works for each.

Tricks to remember Chemical Kinetics

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Zero is Straight, First is Log

  Zero order: [R] against t is a straight line. First order: log[R] against t is a straight line.

  Use it when: Graph-based MCQs and identifying order from data.

• Half-life clue

  If the half-life stays the same as the reaction goes on, it is first order. If it halves as concentration halves, it is zero order.

  Use it when: Case studies with half-life data.

• 2-3-0-3 over t, log of start over now

  Say it aloud: k equals 2.303 over t, log of start over now. 'Start over now' keeps [R]₀ on top.

  Use it when: Writing the first order equation correctly every time.

• Ten degrees, twice the speed

  A 10 °C rise roughly doubles k. 30 °C means 2 × 2 × 2 = 8.

  Use it when: Quick estimates and checking Arrhenius answers.

• Molecularity Must be a whole number

  Molecularity: Must be 1, 2 or 3, never zero. Order: can be anything the experiment says.

  Use it when: Order versus molecularity questions.

Chemical Kinetics: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Chemical Kinetics

The slips examiners see most often, and what to write instead.

• ✗Mistake: Taking order from the balanced equation

  Students assume the coefficients are the powers in the rate law.

  Wrong working

  2NO + O₂ → 2NO₂, so the reaction must be third order without any data.

  Right working

  The order must come from experiment (rate law); it equals the coefficients only for an elementary step.

  ✓Do this: Write 'order is experimental' and use the data given.

  Usually costs: 1 mark in a 2- or 3-mark question.

• ✗Mistake: Inverting the log term in the first order equation

  Wrong working

  k = (2.303/t) log([R]/[R]₀), which gives a negative k

  Right working

  k = (2.303/t) log([R]₀/[R])

  ✓Do this: Initial concentration goes on top; k must come out positive.

  Usually costs: The calculation and answer marks.

• ✗Mistake: Using °C or kJ in the Arrhenius equation

  Temperatures are copied as given and Ea is kept in kJ with R in J.

  Wrong working

  (T₂ − T₁)/(T₁T₂) = 10/(27 × 37)

  Right working

  (T₂ − T₁)/(T₁T₂) = 10/(300 × 310)

  ✓Do this: Convert to kelvin; use R = 8.314 J mol⁻¹ K⁻¹ and Ea in J mol⁻¹, then convert the answer to kJ.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Wrong units for k

  ✓Do this: Use (mol L⁻¹)^(1−n) s⁻¹; for a third order reaction this is L² mol⁻² s⁻¹.

  Usually costs: ½ mark for the unit.

• ✗Mistake: Saying a catalyst shifts the equilibrium

  Students confuse faster with more product.

  ✓Do this: A catalyst speeds up forward and backward reactions equally, so equilibrium is reached faster but K is unchanged.

  Usually costs: The full mark in an MCQ or assertion-reason item.

NCERT exercise guide: Chemical Kinetics

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Chemical Kinetics: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper; 29 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Chemical Kinetics

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Chemical Kinetics question, by year

Practice questions on Chemical Kinetics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which of the following statements is correct?
   (a) Molecularity can be zero.
   (b) Order is always a whole number.
   (c) Molecularity applies only to elementary reactions.
   (d) Order is always equal to molecularity.

   Hint

   Which one is a theoretical idea for a single step?

   Answer

   (c) Molecularity applies only to elementary reactions.

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): A catalyst increases the rate of a reaction.
   Reason (R): A catalyst makes ΔG of the reaction more negative.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   What does a catalyst change, and what does it leave alone?

   Answer

   (c) A catalyst lowers the activation energy; it does not change ΔG.

3. Q3 · 2 marks · Very short answerBoard level

   Define half-life. Derive the expression for the half-life of a first order reaction. (2 marks)

   Hint

   Put [R] = [R]₀/2 into the integrated equation.

   Answer

   Half-life is the time for the reactant concentration to fall to half its initial value (½). k = (2.303/t₁/₂) log 2 (1), so t₁/₂ = 0.693/k (½).

4. Q4 · 3 marks · Short answerHarder board level

   In a zero order reaction the concentration of the reactant falls from 0.50 to 0.30 mol L⁻¹ in 25 minutes. Calculate (a) k, (b) the time for the reaction to go to completion, and (c) the half-life if the initial concentration is doubled to 1.0 mol L⁻¹. (3 marks)

   Hint

   [R] = [R]₀ − kt, and t₁/₂ = [R]₀/2k grows with [R]₀.

   Answer

   (a) k = (0.50 − 0.30)/25 = 8.0 × 10⁻³ mol L⁻¹ min⁻¹ (1). (b) t = 0.50/(8.0 × 10⁻³) = 62.5 min (1). (c) t₁/₂ = 1.0/(2 × 8.0 × 10⁻³) = 62.5 min: doubling [R]₀ doubles the half-life of a zero order reaction (1).

5. Q5 · 3 marks · Short answerBoard level

   Give three differences between order and molecularity of a reaction. (3 marks)

   Hint

   Think: source, possible values, type of reaction.

   Answer

   (1) Order is found experimentally; molecularity is theoretical. (2) Order can be zero or fractional; molecularity is a whole number 1, 2 or 3. (3) Order applies to elementary and complex reactions; molecularity only to elementary steps.

6. Q6 · 3 marks · Short answerHarder board level

   For a reaction with rate law Rate = k[A][B]², what happens to the rate when (a) [B] is doubled, (b) [A] is halved, (c) [A] is tripled and [B] is halved at the same time? (3 marks)

   Hint

   Multiply the factor for each reactant raised to its order.

   Answer

   (a) ×4 (1) (b) ×½ (1) (c) 3 × (½)² = ¾ of the original rate (1).

7. Q7 · 3 marks · Short answerHarder board level

   The half-life of a first order reaction is 30 minutes. (a) What fraction of the reactant remains after 90 minutes? (b) How long will it take for 80% of the reactant to react? (log 5 = 0.699, log 2 = 0.301) (3 marks)

   Hint

   Count half-lives for (a); for (b) 20% is left.

   Answer

   (a) 90 min = 3 half-lives, so ⅛ remains (1). (b) k = 0.693/30 = 0.0231 min⁻¹ (1); t = (2.303/0.0231) log(100/20) = 99.7 × 0.699 ≈ 69.7 min (1).

8. Q8 · 3 marks · Short answerHarder board level

   The reaction 2NO₂ + F₂ → 2NO₂F is thought to go by: NO₂ + F₂ → NO₂F + F (slow); NO₂ + F → NO₂F (fast). (a) Show that the steps add up to the overall equation. (b) Write the rate law and the overall order. (c) Identify the intermediate and the molecularity of the slow step. (d) Why is the reaction not third order even though three molecules appear in the overall equation? (3 marks)

   Hint

   The slow step decides the rate; species made in one step and used in the next do not appear overall.

   Answer

   (a) Adding the steps and cancelling F gives 2NO₂ + F₂ → 2NO₂F (½). (b) Rate = k[NO₂][F₂]; second order (1). (c) F atom; the slow step is bimolecular (½ + ½). (d) Order comes from the rate-determining step (found by experiment), not from the coefficients of the overall equation (½).

9. Q9 · 3 marks · Short answerHOTS / competency

   A plot of log[R] against time for a reaction is a straight line with slope −2.0 × 10⁻³ s⁻¹. (a) What is the order of the reaction? (b) Calculate k and the half-life. (3 marks)

   Hint

   For first order, slope of log[R] against t = −k/2.303.

   Answer

   (a) First order, because log[R] against t is linear (1). (b) k = 2.303 × 2.0 × 10⁻³ = 4.606 × 10⁻³ s⁻¹ (1); t₁/₂ = 0.693/(4.606 × 10⁻³) ≈ 150 s (1).

10. Q10 · 5 marks · Long answerHOTS / competency

    The rate constant of a reaction is 1.0 × 10⁻³ s⁻¹ at 320 K and 1.0 × 10⁻² s⁻¹ at 350 K. (a) Calculate Ea. (b) A catalyst lowers Ea by 11.5 kJ mol⁻¹ at 300 K; by what factor does k change (assume A is unchanged)? (c) Draw and label the energy profile showing both paths. (R = 8.314 J mol⁻¹ K⁻¹) (5 marks)

    Hint

    log 10 = 1; then log(k′/k) = ΔEa/(2.303RT).

    Answer

    (a) 1 = Ea/(2.303 × 8.314) × 30/(320 × 350) (1); Ea = 19.147 × 3733 ≈ 71 500 J mol⁻¹ = 71.5 kJ mol⁻¹ (1). (b) log(k′/k) = 11 500/(2.303 × 8.314 × 300) = 11 500/5744 ≈ 2.0 (1), so k increases about 100 times (1). (c) Reactant and product levels the same for both paths, catalysed hump lower, Ea and Ea′ marked (1).

11. Q11 · 2 marks · Very short answerHOTS / competency

    Why does a small rise in temperature cause a large increase in rate, even though collision frequency rises only slightly? (2 marks)

    Hint

    Think about the tail of the energy distribution.

    Answer

    Rate depends on the fraction of molecules with energy ≥ Ea, given by e^(−Ea/RT) (1). A small rise in T increases this fraction sharply (roughly doubling it for 10 K), so effective collisions increase much more than total collisions (1).

12. Q12 · 1 mark · MCQHarder board level

    For a first order reaction, the ratio of the time for 99.9% completion to the half-life is closest to:
    (a) 2 (b) 3 (c) 7 (d) 10

    Hint

    t₉₉.₉ = (2.303/k) log 1000.

    Answer

    (d) 10, because (2.303 × 3)/0.693 ≈ 9.97.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define instantaneous rate.Show answer

   The rate at a particular moment, given by −d[R]/dt: the slope of the tangent to the concentration–time curve at that time.

2. 2.Give the units of k for zero, first and second order reactions.Show answer

   mol L⁻¹ s⁻¹; s⁻¹; L mol⁻¹ s⁻¹.

3. 3.Can molecularity be zero or fractional?Show answer

   No. Molecularity is always a whole number (1, 2 or 3). Order can be zero or fractional.

4. 4.What is the rate-determining step?Show answer

   The slowest elementary step in a reaction mechanism; it controls the overall rate.

5. 5.Write the half-life of a zero order and a first order reaction.Show answer

   Zero order: [R]₀/2k. First order: 0.693/k.

6. 6.For a first order reaction, how is t₉₉.₉ related to t₁/₂?Show answer

   t₉₉.₉ ≈ 10 × t₁/₂.

7. 7.What does the slope of a plot of ln k against 1/T give?Show answer

   −Ea/R.

8. 8.What two conditions make a collision effective?Show answer

   Enough energy (at least the threshold energy) and proper orientation.

9. 9.Name one zero order reaction.Show answer

   Decomposition of ammonia on a hot platinum surface at high pressure (or of HI on gold).

10. 10.Does a catalyst change ΔG or K?Show answer

    No. It lowers the activation energy only.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write the rate of 2N₂O₅ → 4NO₂ + O₂ in terms of each species?
• ☐Can you find the orders and k from a three-row initial-rate table?
• ☐Can you state three differences between order and molecularity?
• ☐Can you derive [R] = [R]₀ − kt and t₁/₂ = [R]₀/2k?
• ☐Can you find the time for 90% completion of a first order reaction from k?
• ☐Can you use total-pressure data for A(g) → B(g) + C(g) to find k?
• ☐Can you find Ea from rate constants at two temperatures?
• ☐Can you explain the role of energy and orientation in collision theory?

Chemical Kinetics: frequently asked questions